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Hold your nose, it's a Visual C++ 6.5 workaround. For some reason it requires

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boost::hash to define operator() taking const and non-const arguments.


[SVN r32783]
This commit is contained in:
Daniel James
2006-02-09 19:11:54 +00:00
parent 1cf74208aa
commit 514757c312
1 changed files with 57 additions and 7 deletions
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64
include/boost/functional/hash/hash.hpp
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@@ -24,6 +24,10 @@
#include <boost/type_traits/is_array.hpp>
#endif
#if BOOST_WORKAROUND(BOOST_MSVC, < 1300)
#include <boost/type_traits/is_const.hpp>
#endif
namespace boost
{
#if BOOST_WORKAROUND(__BORLANDC__, BOOST_TESTED_AT(0x551))
@@ -310,6 +314,7 @@ namespace boost
// boost::hash
#if !BOOST_WORKAROUND(BOOST_MSVC, < 1300)
template <class T> struct hash
: std::unary_function<T, std::size_t>
{
@@ -323,15 +328,60 @@ namespace boost
{
return hash_detail::call_hash<T>::call(val);
}
#if BOOST_WORKAROUND(BOOST_MSVC, < 1300)
std::size_t operator()(T& val) const
{
return hash_detail::call_hash<T>::call(val);
}
#endif
#endif
};
#else // Visual C++ 6.5
// There's probably a more elegant way to Visual C++ 6.5 to work
// but I don't know what it is.
namespace hash_detail
{
template <class T>
struct hash_impl_base
: std::unary_function<T, std::size_t>
{
std::size_t operator()(T const& val) const
{
return hash_detail::call_hash<T>::call(val);
}
};
template <bool IsConst>
struct hash_impl;
template <>
struct hash_impl<true>
{
template <class T>
struct inner : public hash_impl_base<T> {};
};
template <>
struct hash_impl<false>
{
template <class T>
struct inner : public hash_impl_base<T>
{
std::size_t operator()(T const& val) const
{
return hash_impl_base<T>::operator()(val);
}
std::size_t operator()(T& val) const
{
return hash_impl_base<T>::operator()(val);
}
};
};
}
template <class T> struct hash
: public hash_detail::hash_impl<boost::is_const<T>::value>
::BOOST_NESTED_TEMPLATE inner<T>
{
};
#endif
}
#endif